IA Mathematics SL Volume by revolution

Candidate: 000738-0003

Arandia Jimenez 1

Mathematics SL Internal Assessment

Calculating the Volume of a Designed Flower Vase

Candidate: Maria René Arandia Jiménez Candidate Number: 000738-0003 Session: May 2016 Course: IB Mathematics SL I.

Introduction Calculus is the study of how things change, by which a person can create models and

predict quantitative changes and sequences. (“What Is Calculus and Why do we Study it?”) Differentiation and Integration are the two main branches of Calculus. I was attracted to Integrals; I saw it as a second method of finding the volume of an object, and not only

Candidate: 000738-0003

Arandia Jimenez 2

multiplying the dimensions of a geometrical three-dimensional shape. Considering its purpose, I wondered: Could be possible to calculate the volume of non-regular shapes, such as a vase? Figure 1. shows a sine function on a graph with its shaded area under the line, and Figure 2. shows how it looks when the area is rotated 3600 around the x-axis.

Figure 1. Sketch of a sine function on a set of axis

Figure 2. Sketch of a rotated sine functions

on a set of axis

on a set of axis

Figure 1. and 2; “Solid of Revolution - Finding Volume by Rotation”; WyzAnt; wyzant.com; Web; 10 March, 2016. r is changing for every point in the x-axis, and if r ¿ f ( x )= y , the area of the imaginary cross sectional circle inside the vase will also be changing, because it is an irregular 3D shape. In 2 order to find the volume of the vase using the area ( A=π r ¿ of its imaginary cross sectional

disks inside, the volume of a cross sectional disk has to be calculated. The area of the disk should be multiplied by its infinitesimally small thickness (dx). V d =π r 2 dx Next, we have to integrate in order to find the volume of the entire base. Integration will add up the volume of all disks that form the vase, considering that the volume of each is changing for all values of x. The area under the function f(x) will be rotated around the x-axis, with limits established by vertical lines x=a and x=b. (“Clip 2: Solids of Revolution”). a

V=

∫πr b

a 2

dx

→

V=

∫ π ( f (x))2 dx b

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Arandia Jimenez 3

I felt enthusiastic to find the volume of an object, but to challenge myself, I decided to create a flower vase and model it. While designing the vase, I got inspired by the Golden ratio, which are used for artworks and architecture since Ancient Greece, and is found in many aspects of nature, such as flowers. A Golden spiral created by the Golden rectangle will be fancy! 3. Math ImageWorld; of the mathworld.wolfram.com; Golden spiral Figure 3; “Golden Spiral”;Figure Wolfram Web; 15 March,

2016. The Golden ratio φ

is to be obtained through the division of length (L) of the rectangle

by its width (x): Golden ratio=φ ≈ The specific value of φ

L W

is achieved by getting a 2 by 2 square, and finding a midpoint

on its base, where a diagonal line will be drawn until it touches the upper right corner of the square. Then, draw a curve-like line from the upper right corner to the bottom of the square, and connect it with a horizontal line to the midpoint mentioned before. Draw the margins of the new rectangle. The final drawing is shown in Figure 4: Figure 4. Drawing of how to find φ

φ=

L W

Candidate: 000738-0003

φ=

1+ √ 5 2

Arandia Jimenez 4

(using a 2x2 cm square) ≈ 1.6180339887 … .

L will be calculated by adding half on the square’s base and the length of the red diagonal:

√

x x 2 L= + x 2+( ) 2 2 I will use this method to find the distance between the points of the functions that follow the golden spiral shape, so I can model the vase and then find a best-fit function that will allow me to find its volume. Finally, I will rotate the area to find the volume. II.

Aims

Integrals give us humans a method to find the volume of original and non-conventional objects. Therefore, the aim of this exploration is to apply integrals in a possible real life situation, in which I created and modeled a decorative and irregular solid object. With Paint program, I could portray the idea that I had of creating a flower base, inspired in the golden spiral, as seen in Figure 5. Figure 5.Contour and dimensions of the created flower vase.

Figure 5. is the contour of the vase: the black lines represent the outer most surface of the vase, and red vase are the inside contour. The volume between the lines will be calculated.

Candidate: 000738-0003 III.

Arandia Jimenez 5

Modeling In Figure 6. I had to define the points that I wanted to graph. Finding the points will be

easy for both of the lines (before the yellow dots), but from the yellow dots to the right, the lines follow the golden ratio proportions, so I had to decide what dimensions the square will have, in order to find the other points on the golden spiral. Figure 6. Coordinate plane and draft sketch of the functions, with defined square dimension

I decided that the dimensions of the squares will be 4x4cm for the red spiral (ABCD) and 2.5x 2.5cm for the black spiral (EFGH). I will refer to the length as the longer side of the rectangle, and width as the shorter side. Red Spiral (ABCD) As shown in Figure 6, the green points have to be located on the coordinate plane. Point A on the red spiral is already defined as (14.9, 4.7), so it is necessary to find the point B. It is obvious that point B will be at (18.9, 8.7) because of the established dimensions of the square (4cm to the right and 4cm upwards). To find point C, the length of the entire rectangle has to be

calculated with the equation found before:

√

x x 2 L= + x 2+( ) 2 2

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The width of the 1st square (

DS 1

) is 4, as shown in Figure 6, and it will replace x in the

equation to find the length of the first rectangle (

LR 1

):

√

4 4 2 LR 1= + 4 2+( ) 2 2 LR 1=2+2 √ 5

≈ 6.4721

The answer is given in 4 decimal places. To find the y-coordinate of point C ( y-coordinate of point A (

yA

yC

), the

) has to be added by the length of the 1st rectangle y C = y A + LR 1 y C =4.7+(2+ 2 √ 5) ≈11.1721

To find the x-coordinate of point C, the dimensions of the 2nd square (

DS 2

) have to be

found, DS 2=L R 1−4 D S 2=2+ 2 √ 5−4 ≈ 2.472 and then subtract the answer from the x-coordinate of point B. x C =18.9−(2+2 √5−4 ) ≈ 16.428 Therefore, point C has coordinates (16.428, 11.172). The 1st rectangle can be divided to form a 2nd square and a 2nd rectangle. The 2nd square will have the dimensions 2.472x2.472 cm as shown in Figure 7. Figure 7. Dimensions

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The length of the 2nd rectangle is 4cm, but I wanted to make sure that my numbers were accurate, so I decided to make the same process this time replacing the dimensions of the 2nd square ( 2+2 √ 5 – 4 or ≈2.472) in the formula, and expecting a 4 as an answer.

√

x x 2 LR 2 = + x 2 +( ) 2 2

LR 2 =

√

2

2 2+2 √ 5 – 4 2+2 √5 – 4 + (2+ 2 √5 – 4) +( ) 2 2

LR 2 =4 This means that the process is correct. The same process can be repeated to find point D, but, since I have enough information, I will just find the dimensions of the 3rd square by subtracting the length of the 2nd rectangle by one side of the 2nd square: DS 3 =LR 2 −D S 2 2+2 √5 – 4 D S 3=4−¿ )

D S 3 =6−2 √ 5 ≈ 1.528

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Therefore, the coordinates of point D will be obtained by: y D = y C −D S 3 6−2 √5 y D =( 6.7+2 √ 5)−¿ ) y D =0.7+ 4 √ 5 ≈ 9.644 The x-coordinate of point D is the same as point A’s. Point D has coordinates (14.9, 9.644) The points found on the red spiral (ABCD) were:    

A(14.9, 4.7) B(18.9, 8.7) C(16.428, 11.172) D(14.9, 9.644)

To proof if the points on the Golden spiral were found correctly, I have test the dimensions of the rectangles followed the Golden ratio: φ=

L 1+ √ 5 = ≈ 1.6180339887 … . W 2

Table 1. Dimensions of Red rectangle and proof of Golden ratio in Red spiral L W

Rectangle

L (cm)

W (cm)

1R

2+2 √5

4

1+ √ 5 2

2R

4

2 √ 5−2

1+ √ 5 2

3R

2 √ 5−2

6−2 √ 5

1+ √ 5 2

Proof:

φ=

Candidate: 000738-0003

Arandia Jimenez 9 LR1 =φ W R1 4 1+ 5 = √ 2 2 √ 5−2 2 5+1 1+ √5 ∙( √ )= 2 √5−1 √ 5+1 2 √ 5+2 1+ √ 5 √ 5+1 1+ √ 5 = = 4 2 2 2

▄ Black Spiral (EFGH) I used the exact same procedure to find the coordinates of the black spiral (EFGH). The points found on the black spiral (EFGH) were:    

E(15.6, 6.6) F(18.1, 9.1) G(16.555, 10.645) H(15.6, 9.690)

The following table shows the dimensions of the rectangles and how they proved to follow the Golden ratio. Table 2. Dimensions of Black rectangle and proof of Golden ratio in Black spiral L W

Rectangles

L (cm)

W (cm)

1B

5+ 5 √ 5 4

2.5

1+ √ 5 2

2B

2.5

−5+5 √ 5 4

1+ √ 5 2

φ=

Candidate: 000738-0003 3B

Arandia Jimenez 10 −5+5 √ 5 4

15−5 √ 5 4

Proof: LR1 =φ W R1 −5+5 √ 5 ) 4 1+ 5 = √ 2 15−5 √ 5 ( ) 4 (

−1+ √ 5 3+ √ 5 1+ √ 5 ∙( )= 2 3− √ 5 3+ √ 5 2 √ 5+2 1+ √ 5 √ 5+1 1+ √ 5 = = 4 2 2 2

▄

Figure 8. Graph showing set of functions and its coordinate points

1+ √ 5 2

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The following set of data shows three functions for the black shape and its points: Table 3. Points for the functions in Black shape. x [cm] f(x) [cm] x [cm] g(x) [cm] x [cm] h(x) [cm]

0 4.2 15.6 6.6 18.1 9.1

4.9 8 16.3 7 17.2 10.3

7.9 8.2 17 7.5 16.555 10.645

12.5 7.1 17.6 8.1 15.9 10.2

15.6 6.6 18.1 9.1 15.6 9.690

The following set of data shows three functions for the red shape and its points: Table 4. Points for the functions in Red shape x[cm] i(x) [cm] x[cm] j(x) [cm] x[cm] k(x) [cm]

1.4 0 14.9 4.7 18.9 8.7

2.5 4 15.9 5.1 18.2 10.3

7.2 6.3 17.4 5.9 16.428 11.172

11.2 5.3 18.6 7.1 15.3 10.8

14.9 4.7 18.9 8.7 14.9 9.644

Black Shape Function 1 (f(x)) For function f(x), a cubic model will be used: y=ax3 +b x 2 +cx +d To find variables a, b, c, and d, I will replace points (4.9, 8), (7.9, 8.2), (12.5, 7.1), and (15.6, 6.6) in the equation, and create a system (

S1

) of 4 equations with 4 variables:

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Arandia Jimenez 12

 ( 4 .9 ) 3 a  ( 4 .9 ) 2 b  4 .9 c  d  8







3 2  ( 7 .9 ) a  ( 7 .9 ) b  7 .9 c  d  8 .2    3 2  (12.5) a  (12.5) b  12.5c  d  7.1 

S1 =

 (15.6) 3 a  (15.6) 2 b  15.6c  d  6.6  

which will become:  117.649a  24.01b  4.9c  d  8 1  493.039a  62.41b  7.9c  d  8.2 2   S 1=¿  1953.125a  156.256b  12.5c  d  7.1  3    3796.416a  243.366b  15.6c  d  6.6 4

I will solve the system of equations by elimination: Subtract 1 from 2 to eliminate d: ¿ 493.039 a+62.41 b+7.9 c+ d=8.2 117.649 a+ 24.01b+ 4.9 c +d=8 375.39 a+38.4 b+ 3 c=0.2 A Subtract 3 from 4 to eliminate d: ¿ 3796.416 a+ 243.366 b+15.6 c +d=6.6 1953.125 a+156.256 b+ 12.5 c+ d=7.1 1843.291a+ 87.11b+3.1 c=−0.5 B Subtract 2 from 3 to eliminate d: ¿ 1953.125 a+156.256 b+ 12.5 c+ d=7.1 493.039 a+62.41 b+7.9 c +d=8.2 1460.086 a+93.846 b +4.6 c=−1.1 C Subtract A from B to eliminate c:

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Arandia Jimenez 13

¿ 1843.291a+ 87.11 b+3.1 c=−0.5(¿ 3) 375.39 a+38.4 b+3 c =0.2(¿ 3.1) ¿5529.873 a+ 261.33b+ 9.3 c=−1.5 1163.709 a+ 119.04 b+ 9.3 c=0.62 4366.164 a+142.296 b=−2.12 I Subtract A from C to eliminate c: ¿ 1460.086 a+93.846 b+ 4.6 c=−1.1(¿ 3) 375.39 a+38.4 b +3 c=0.2(¿ 4.6) ¿ 4380.258 a+281.538 b+13.8 c=−3.3 1726.794 a+176.64 b+ 13.8 c=0.92 2653.464 a+104.898 b=−4.22 II Subtract I from II to eliminate b: ¿ 2653.464 a+104.898 b=−4.22(¿ 142.296) 4366.164 a+142.296 b=−2.12(¿ 104.898) ¿ 377577.3133 a+14926.56581 b=−600.48912 458001.8713 a+14926.56581 b=−222.38376 −80424.558 a=−378.10536 II

a=

−378.10536 −80424.558

a ≈ 4.7014 * 10−3

Replace a in I to find b: 4.7014 −3 4366.164 ¿ * 10 ¿+142.296 b=−2.12 b ≈−0.1592

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Arandia Jimenez 14

Replace a and b in A to find c: 4.7014 −3 10 ¿+38.4(−0.1592)+ 3 c=0.2 375.39 ¿ * c ≈1.5161 Replace a, b, and c in 1 4.7014 −3 10 ¿+24.01(−0.1592)+ 4.9(1.5161)+d =8 117.649 ¿ * d ≈ 3.8404

Function f(x) results in: y=4.7014 * 10−3 x 3−0.1592 x 2 +1.5161 x +3.8404

Equation test: (4.9, 8): Replace x = 4.9 on f(x): y=4.7014 * 10−3 (4.9)3−0.1592 ( 4.9 )2+1.5161( 4.9)+ 3.8404

y ≈ 8.0000 (7.9, 8.2): Replace x = 7.9 on f(x): y=4.7014 * 10−3 (7.9)3−0.1592 ( 7.9 )2+1.5161(7.9)+3.8404 y ≈ 8.1999

(15.6, 6.6): Replace x = 15.6 on f(x):

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Arandia Jimenez 15

y=4.7014 * 10−3 (15.6)3−0.1592 ( 15.6 )2 +1.5161(15.6)+ 3.8404

y ≈6.5971

Figure 9. Graph of manual and automatic curve fit of f(x).

The auto fit curve on Logger Pro has an equation of: 3

2

f (x)=0.004314 x −0.1469 x +1.395 x+ 4.198 After the equation test, I graphed the points on Logger Pro package program (pink line) and could see that the auto fit curve and the manual fit (blue line) curve were very close to each other. Since the pink curve passes through all the points of the function f(x), I decided that the automatic curve fit will be used in order have a more precise results. Function 2 (g(x)) For function g(x), a cubic model will be used: y=ax3 +b x 2 +cx +d

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Arandia Jimenez 16

To find variables a, b, c, and d, I will replace points (16.3, 7), (17, 7.5), (17.6, 8.1), and (18.1, 9.1) in the equation, and create a system (

S2

) of 4 equations with 4 variables:

 (16.3) 3 a  (16.3) 2 b  16.3c  d  7   S2 =

 3 2  (17) a  (17) b  17c  d  7.5    3 2  (17.6) a  (17.6) b  17.6c  d  8.1  (18.1) 3 a  (18.1) 2 b  18.1c  d  9.1   

which will become:

S2 =

 4330.747 a  265.69b  16.3c  d  7   4913a  289b  17c  d  7.5       5451.776a  309.76b  17.6c  d  8.1  5929.741a  327.64b  18.1c  d  9.1 

Using a Casio fx-9860GII SD graphics calculator, I found the 4 variables, rounded to 4 decimal places: A=0.3829 B=−19.2724

C=323.9507 D=−1811.3755

and replaced the variables in the cubic equation, and the resulting function g(x) is: y=0.3829 x 3−19.2724 x 2 +323.9507 x−1811.3755 Equation test: (16.3, 7): Replace x = 16.3 on g(x): y=0.3829( 16.3)3 −19.2724 ( 16.3 )2+323.9507 (16.3)−1811.3755

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Arandia Jimenez 17

y ≈6.7800 (17, 7.5): Replace x = 17 on g(x): y=0.3829(17)3 −19.2724 ( 17 )2 +323.9507(17)−1811.3755 y=7.2505

(17.6, 8.1): Replace x = 17.6 on g(x): y=0.3829(17.6)3 −19.2724 ( 17.6 )2 +323.9507(17.6)−1811.3755 y=7.8232

Figure 10. Graph of manual and automatic curve fit of g(x)

The auto fit curve on Logger Pro has an equation of: g( x)=0.1866 x 3−9.113 x 2 +148.9 x −806.9

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After the equation test, I graphed the points on Logger Pro package program (pink line) and observed that the auto fit curve was more accurate than the manual fit curve (blue line) because it passed through all the points of the function g(x). Again, I decided that the automatic curve fit will be used in order have a more precise calculation of the volume. Function 3 (h(x)) For function h(x), a cubic model will be used: y=ax3 +b x 2 +cx +d To find variables a, b, c, and d, I will replace points (17.2, 10.3), (16.555, 10.645), (15.9, 10.2), and (15.6, 9.690) in the equation, and create a system (

S3

) of 4 equations with 4

variables:  (17.2) 3 a  (17.2) 2 b  17.2c  d  10.3 

 

(16.555) 3 a  (16.555) 2 b  16.555c  d  10.645 S 3  =  3 2  (15.9) a  (15.9) b  15.9c  d  10.2   (15.6) 3 a  (15.6) 2 b  15.6c  d  9.690    which will become:  5088.448a  295.84b  17.2  d  10.3   4537.19612a  274.068b  16.555c  d  10.645    S 3=¿  4019.679a  252.81b  15.9c  d  10.2    3796.416a  243.36b  15.6c  d  9.690 

Using a Casio fx-9860GII SD graphics calculator, I found the 4 variables, rounded to 4 decimal places: A=0.0842 B=−5.1127 C=100.12 24

D=−627.4719

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Arandia Jimenez 19

and replaced the variables in the cubic equation, and the resulting function h(x) is: y=0.0842 x 3−5.1127 x 2 +100.1224 x−627.4719 Equation test: (17.2, 10.3): Replace x = 17.2 on h(x): y=0.0842(17.2)3−5.1127 (17.2 )2 +100.1224 (17.2)−627.4719 y ≈10.5395 (16.555, 10.645): Replace x = 16.555 on h(x): y=0.0842(16.555)3−5.1127 ( 16.555 )2+100.1224 (16.555)−627.4719 y ≈10.8588

(15.9, 9.690): Replace x = 15.9on h(x): y=0.0842(15.9)3−5.1127 ( 15.9 )2+100.1224 (15.9)−627.4719 y ≈10.3895

Figure 11. Graph of manual and automatic curve fit of h(x).

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The auto fit curve on Logger Pro has an equation of: h( x )=0.1630 x 3−8.989 x 2+163.6 x−973.5 With Logger Pro package program I observed that, again, the auto fit curve (pink line) was more accurate than the manual fit curve (blue line) because it passes through all the points of the function h(x). Again, my choice was to use the equation given by Logger Pro, as more accurate results will be obtained. Red Shape Function 4 (i(x)) The functions calculated were close, but not very accurate as the variables found by Logger Pro. From here on, the variables of the functions i(x), j(x), and k(x) are to be calculated through Logger Pro’s automatic curve fit. For function i(x), a cubic model will be used, taking the form of: 3

2

y=ax +b x +cx +d

Figure 12. Graph of manual and automatic curve fit of i(x).

Candidate: 000738-0003

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Based on what the graph provides, i(x) can be: i ( x )=0.01573 x 3−0.4696 x 2 +4.120 x−4.382 Function 5 (j(x)) Figure 13. Graph of manual and automatic curve fit of j(x).

For function j(x), a cubic model will be used, taking the form of: j ( x )=0.1551 x 3 −7.585 x 2 +123.8 x −669.1 Function 6 (k(x)) Figure 14. Graph of manual and automatic curve fit of k(x).

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A cubic model will be used for function k(x), and with Logger Pro model, it will result: k ( x )=−0.005484 x 3−0.2414 x 2 +12.61 x−106.3 Two systems of equations can by constructed by assembling the functions that make up each shape: Black shape system (

IV.

) and Red shape system (

{

SR

)

f ( x ) =0.004314 x 3−0.1469 x 2+1.395 x +4.198 ; for 0 ≤ x ≤15.6 ¿ g ( x )=0.1866 x 3−9.113 x 2 +148.9 x−806.9 ; for 15.6 ≤ x ≤ 18.1 h ( x )=0.1630 x 3−8.989 x 2 +163.6 x −973.5 ; for 15.6 ≤ x ≤ 18.1

SB

SR

SB

{

}

i ( x )=0.01573 x3 −0.4696 x2 + 4.120 x −4.382 ; for 1.4 ≤ x ≤ 4.9 ¿ j ( x )=0.1551 x 3−7.585 x2 +123.8 x−669.1 ; for 14.9 ≤ x ≤ 18.9 k ( x )=−0.005484 x 3−0.2414 x2 +12.61 x−106.3 ; for 14.9 ≤ x ≤ 18.9

}

Integration

The next step is to find the volume through integrals. This process will have 3 stages. I will integrate the functions found by using the equation: a

V=

∫ π ( f (x))2 dx b

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Stage 1: Figure 15. Volume 1 (

V1

) (shaded yellow area) that is to be found in Stage 1

Replace f(x) in equation: a

V f (x)

=

∫ π (0.004314 x 3−0.1469 x 2 +1.395 x + 4.198)2 dx b

V f (x)

15.6

=

6

∫ π ( ( 1.8611∗10−5 ) x −(1.2675∗10−3 ) x 5 +0.0336 x 4−0.3736 x 3 +0.7126 x 2+11.7124 x+ 17.6232)dx 0

[

V f (x) = π (

1.8611∗10 7

−5

7

x −

−3

1.2675∗10 6

6

x+

V f (x) ≈ 2613.7947 cm3 For the following results of the integrated functions, Casio fx-9860GII SD graphics calculator was used to find the volume, with 4 decimal places. Replace g(x) in equation:

]

0.0336 5 0.3736 4 0.7126 3 11.7124 2 15.6 x− x + x + x +17.6232 x) 5 4 3 2 0

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Arandia Jimenez 24 18.1

V g (x)

∫ π (0.1866 x 3−9.113 x 2 +148.9 x−806.9)2 dx

=

15.6

V g (x) ≈ 452.1702 cm3 Replace i(x) in equation: 14.9

V i( x)

=

∫ π (0.01573 x3 −0.4696 x 2+ 4.120 x −4.382)2 dx 1.4

V i( x) ≈ 1239.1658 cm3 Replace j(x) in equation (limit: a = 18.1): 18.1

V j (x)

=

∫ π (0.1551 x 3−7.585 x 2 +123.8 x −669.1)2 dx

14.9

V j (x) ≈ 293.7897 cm3

Add Black shape volumes and Red shape volumes, separately. Then subtract to find

V1

V ¿ =V f (x) +V g(x) V ¿ =V i( x)+ V j (x) V ¿ =2613.7947+452.1702=3065.9649V ¿ =1239.1658+293.7897 =1532.9555

V 1=V ¿ −V ¿ V 1=1533.0094 cm3 Stage 2: Figure 16. Volume 2 (

V2

) (shaded yellow area) that is to be found in Stage 2.

:

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Arandia Jimenez 25

Replace k(x) in equation: 18.9

V k (x)

=

∫ π (−0.005484 x3 −0.2414 x 2+ 12.61 x−106.3)2 dx

14.9

V k (x) ≈ 1444.6748 cm3 Replace h(x) in equation: 18.1

V h( x)

=

∫ π (0.1630 x 3−8.989 x 2+163.6 x−973.5)2 dx

15.6

V h( x) ≈ 850.5552 cm3

Subtract to find

V2

:

V 2=V k ( x )−V h (x ) 1444.6748−850.5552 V 2=594.1196 cm3 As shown in Figure 10, there are two pink shaded areas (I and II) that are not subtracted from

V k ( x)

, so

V2

will not be accurate. To fix this, I found both of the functions: I(x) from

points D to H, and II(x) from points F to B. I replaced them in the integrals equation:

Candidate: 000738-0003 I ( x )=

Arandia Jimenez 26

23 −x x−2.27 II ( x )= +18.15 30 2 15.6

V I (x) = ∫ π ( 14.9

18.9

2 2 23 −x x−2.27) dx V II ( x)= ∫ π ( + 18.15) dx 30 2 18.1

V I (x) =195.2634 cm3 V II (x)=199.1200 cm3 V Final 2 =V 2−(V I ( x ) +V II ( x ) ) V Final 2 =199.7362cm

3

Stage 3: Figure 17. Volume 3 (

V3

) (shaded yellow area) that is to be found in Stage 3

Replace j(x) ( 18.1≤ x ≤ 18.9 ) in equation: 18.9

V j (x)

=

∫ π (0.1551 x 3−7.585 x 2 +123.8 x −669.1)2 dx

18.1

V j (x) ≈ 135.5808 cm3

Subtract

V j (x)

from

V II (x)

to get Volume 3:

Candidate: 000738-0003

Arandia Jimenez 27 V 3=V II ( x )−V j(x ) V 3=63.5392 cm3

The three volumes found have to be added to find the volume of the entire flower vase: V =V 1 +V 2 +V 3 V =1533.0094+ 199.7362+ 63.5392 V =1796.2848 cm 3 V.

Conclusion and Reflection

Throughout the process of this exploration I was able to learn new mathematical concepts, as well as apply previous knowledge on a real life situation. The new skill I learned was how to find points on a graph that follow the Golden spiral and the Golden ratio. Not only have I practiced finding the volume of 3D solid objects by integrals, but I also improved my skills on finding the volume of irregular solid objects that require integrating more than 3 functions to be solved. For me, this mathematical exploration was a proof that the disc method on a solid of revolution can be applied in a real life situation or possible real life situation. This means that this method can be used to find the volume of several, and different, irregular solid objects, such as bottles and even more types of flower vases. There are so many objects that can be modeled to find its volume that we, humans, will never get tired of applying calculus in several and unique forms! My aim of finding the volume of a non-conventional object was fulfilled. After achieving my aims, I was thinking where else I could apply the processes of modeling and integrating to find volume of this flower vase; I wanted to find a real life situation where this would be useful. The first idea that came to my mind was that I could use it to optimize the price of it. This means that I could figure out what is the least or greatest amount of material (glass for example) that could be used to create this “proposed” flower vase, and then find the cost of production by finding the price of the material per

cm3 , and multiply it by the volume calculated. This process is done

Candidate: 000738-0003

Arandia Jimenez 28

by several industries, which gives us another reason to proof that calculus is very important in real life. VI. References  Clip 2: Solids of Revolution. Session 57: How to Calculate Volumes. Massachusetts 

Institute of Technology, 2006. Web. 10 Mar. 2016. http://ocw.mit.edu/ Kleitman. "What Is Calculus and Why Do We Study It?" Professor Kleitman's



Homepage. N.p., n.d. Web. 9 Mar. 2016. . Solid of Revolution - Finding Volume by Rotation. Digital image. WyzAnt. N.p., n.d.



Web. 10 Mar. 2016. . Weisstein, Eric W. Golden Spiral. Digital image. Wolfram Math World. N.p., n.d. Web. 10 Mar. 2016. .

Programs and tools used in the exploration:    

Casio fx-9860GII SD graphics calculator Microsoft Word Logger Pro graphing software Paint graphics program